Work done against gravity, what am i missing

[lignocaine]

Hi everyone.

Im studying basic physics for my anaesthesia primary exams and came across the subject of work. I am however finding some difficulty grasping the concept of work done against gravity. The example given in my literature for work done against gravity is the following:

lifting an apple of 102grams 1m vertically will require 1joule of energy.

here it is explained that because the apple exerts a downward force of 1N (0.102kg x 9.8m/s) that an equally opposite force of 1N will need to be exerted upwards to move the object against gravity.

But, if 1N is exerted in an upward direction, the apple will remain still in your hand because the net forces acting on the apple are zero.

So in order to move the apple against gravity, a force of greater than 1N will have to be exerted by your hand.
An example would be using a force of 1.5N over a distance of 1m which would effectively yield a force of 0.5N acting upwards and if this example is used, the work done to move the apple would be 0.5N x 1m = 05J
If a higher upward acting force was used, then the resulting work done to move the object 1m would be even higher

So, the work done to move a fixed mass object 1m against gravity can vary depending on the net force acting on the object?

Is my logic flawed and if yes, where?

 

[arildno]

Do not mix together acceleration and velocity!

Balance of forces (i.e, net force equalling zero!) means zero acceleration.
But you can perfectly well have zero acceleration and a non-zero, constant velocity.

Besides, when we think of work done BY a force, we are interested in finding that particular force's contribution to the total work done.

Therefore, it is wrong to use the "net force" in order to calculate a particular force's work contribution.

 

[lignocaine]

I agree, but here's an example of the same object, the apple starting at rest.
initial velocity = 0m/s
force of gravity balanced by force of hand against apple. In this case there is no acceleration and therefore net force acting on the apple is zero.
If any additional upward force were to act on the apple, it would accelerate upwards (mass x acceleration) if that force acted over a distance of d vertical height, then the work done do move the object from rest upwards against gravity for a distance of d is net force F(net) X d

Again here i am inclined to think that the force used against the downward force of gravity will determine how much work is done to move an object a set distance upwards.

I haven't done physics for a long time and now i think I am confusing myself!

 

[Doc Al]

You are asking to determine the work that you need to do to raise the apple, not the net work on the apple. To raise the apple, you must exert a force equal to its weight (maybe a touch more) over a distance "d": Thus you do work on the apple. The work you do is what is being asked for. Of course, the net work on the apple is zero: It starts at rest and ends at rest.

 

[Andy Resnick]

Also, don't confuse force and work. Holding the apple up in a static position requires force but no work is performed. Moving the lifted apple horizontally also requires no force or work (try telling that to your muscles...). The work you performed is equal to a force times a distance.

The amount of work required to lift the apple is m*g*h, regardless of the amount of time it took to raise the apple.

 

[lignocaine]

Im still a little confused.
How would you answer this then:

how much work is required to move an apple with a mass of 102g over a vertical distance of 1m?

Is the work dependent on how much force u apply or will the work done to move the apple always be constant?

I keep thinking that because the force i can apply to the apple can be variable, the work done would also be variable.

 

[Doc Al]

Im still a little confused.
How would you answer this then:

how much work is required to move an apple with a mass of 102g over a vertical distance of 1m?

The minimum work required to overcome gravity is mg*d = 0.102*9.8*1 = 1 J.

Is the work dependent on how much force u apply or will the work done to move the apple always be constant?

I keep thinking that because the force i can apply to the apple can be variable, the work done would also be variable.

You are correct. You can push as hard as you want on that apple and thus do even more work than the minimum. You'll end up moving the apple up one meter and increasing its speed.

Generally, when they ask about the work needed to lift something they mean the minimum work needed to just lift it without giving it any added speed (and thus added kinetic energy).

Good questions!

 

[polar]

Im still a little confused.
How would you answer this then:

how much work is required to move an apple with a mass of 102g over a vertical distance of 1m?

Is the work dependent on how much force u apply or will the work done to move the apple always be constant?

I keep thinking that because the force i can apply to the apple can be variable, the work done would also be variable.

If the apple is stationary in your hand when you begin, and then also stationary 1m away when you are finished, then the exact same amount of work must have been used to accelerate the apple upwards as was used to decelerate the apple (accelerate it in the opposite direction) once it reaches its destination.

Although it may be difficult to visualize, there is no real difference between that action and the act of accelerating and then braking in a car, except that one is in a vertical direction and the other is in a horizontal direction.

Once you can accept that those two forces (acceleration at the start/deceleration at the end) - and hence the work required for acceleration/deceleration - exactly counteract each other, then you can focus on the remaining part of the question which is how much net work was required to actually lift the apple.

 

[CausticPotash]

The minimum work required to overcome gravity is mg*d = 0.102*9.8*1 = 1 J.


You are correct. You can push as hard as you want on that apple and thus do even more work than the minimum. You'll end up moving the apple up one meter and increasing its speed.

Generally, when they ask about the work needed to lift something they mean the minimum work needed to just lift it without giving it any added speed (and thus added kinetic energy).

Good questions!

I am wondering whether the minimum work required to overcome gravity which is 1 J here can only lift the apple stationary in ur hand without being able to move up or down?
If u want to move it up more, then u will have to use more force (also more work done, of course) than 1 J? Am I right?
Moreover, in any case, the force on an object by gravity is the object's weight itself (in this case, the apple's weight)? is it a negative force?

 

[Lsos]

To raise the apple 1m requires 1J of work.

You are right in that you might need to use a bit more than 1N of force to get the thing moving, but that means that you also accelerated it, and can therefore stop pushing a bit before the apple is 1m high...and let its momentum take it there. However you look at it, it ends up gaining 1J of energy.

Also, I think you are confusing energy and force. To overcome gravity, you must push the apple with 1N. But, just holding it in place with 1N is not doing work. You're doing nothing. You can substitute a table for your hand, and the table will supply that 1N indefinitely. Hell, you can ask the table to support a car with 10,000N...and it's still no work is being done. Only by pushing it with 1N AND moving it 1m, will you get 1J. Without that motion against the force, you get nothing.

Moreover, in any case, the force on an object by gravity is the object's weight itself (in this case, the apple's weight)? is it a negative force?

By saying a force is "negative", you're just talking about which direction that force is pushing/ pulling. If we say that up is "positive", then "down" (gravity), would be negative. But, we can arbitrarily set whichever direction we want to be "positive" or "negative", so I could just as well say that up is negative, thereby making gravity positive...

 

[Philip Harris]

Thanks for explaining this